Error when converting from pyspark RDD to DataFrame: Cannot infer schema of type 'unicode' [duplicate] - apache-spark

Question

Could someone help me solve this problem I have with Spark DataFrame?
When I do myFloatRDD.toDF() I get an error:
TypeError: Can not infer schema for type: type 'float'
I don't understand why...
Example:
myFloatRdd = sc.parallelize([1.0,2.0,3.0])
df = myFloatRdd.toDF()
Thanks

Answer 1

SparkSession.createDataFrame, which is used under the hood, requires an RDD / list of Row/tuple/list/dict* or pandas.DataFrame, unless schema with DataType is provided. Try to convert float to tuple like this:
myFloatRdd.map(lambda x: (x, )).toDF()
or even better:
from pyspark.sql import Row
row = Row("val") # Or some other column name
myFloatRdd.map(row).toDF()
To create a DataFrame from a list of scalars you'll have to use SparkSession.createDataFrame directly and provide a schema***:
from pyspark.sql.types import FloatType
df = spark.createDataFrame([1.0, 2.0, 3.0], FloatType())
df.show()
## +-----+
## |value|
## +-----+
## | 1.0|
## | 2.0|
## | 3.0|
## +-----+
but for a simple range it would be better to use SparkSession.range:
from pyspark.sql.functions import col
spark.range(1, 4).select(col("id").cast("double"))
* No longer supported.
** Spark SQL also provides a limited support for schema inference on Python objects exposing __dict__.
*** Supported only in Spark 2.0 or later.

Answer 2

from pyspark.sql.types import IntegerType, Row
mylist = [1, 2, 3, 4, None ]
l = map(lambda x : Row(x), mylist)
# notice the parens after the type name
df=spark.createDataFrame(l,["id"])
df.where(df.id.isNull() == False).show()
Basiclly, you need to init your int into Row(), then we can use the schema

Answer 3

Inferring the Schema Using Reflection
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext
# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to Row
orders_struct = parts.map(lambda p: Row(order_id=int(p[0]), order_date=p[1], customer_id=p[2], order_status=p[3]))
for i in orders_struct.take(5): print(i)
#convert the RDD to DataFrame
orders_df = spark.createDataFrame(orders_struct)
Programmatically Specifying the Schema
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext
# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to tuple
orders_struct = parts.map(lambda p: (p[0], p[1], p[2], p[3].strip()))
#convert the RDD to DataFrame
orders_df = spark.createDataFrame(orders_struct)
# The schema is encoded in a string.
schemaString = "order_id order_date customer_id status"
fields = [StructField(field_name, StringType(), True) for field_name in schemaString.split()]
schema = Struct
ordersDf = spark.createDataFrame(orders_struct, schema)
Type(fields)

Answer 4

from pyspark.sql import Row
myFloatRdd.map(lambda x: Row(x)).toDF()